The standard patch for a torus is given by \begin{align} \xx(\phi,\psi) &= (R+\rho\cos\psi)\cos\phi\,\xhat + (R+\rho\cos\psi)\sin\phi\,\yhat + \rho\sin\psi\,\zhat \end{align} so that the coordinate basis for tangent vectors is \begin{align} \ev_1 &= \frac{\partial\xx}{\partial\phi} = (R+\rho\cos\psi) (-\sin\phi\,\xhat + \cos\phi\,\yhat) ,\\ \ev_2 &= \frac{\partial\xx}{\partial\psi} = -\rho\sin\psi\cos\phi\,\xhat - \rho\sin\psi\sin\phi\,\yhat + \rho\cos\psi\,\zhat , \end{align} and the normal vector is \begin{align} \nv &= \ev_1\times\ev_2 = \rho(R+\rho\cos\psi) (\cos\psi\cos\phi\,\xhat + \cos\psi\sin\phi\,\yhat + \sin\psi\,\zhat) \\ \end{align} which when normalized becomes \begin{align} \nn &= \cos\psi\cos\phi\,\xhat + \cos\psi\sin\phi\,\yhat + \sin\psi\,\zhat . \end{align} We can now determine the shape operator by taking derivatives of $\nn$, yielding \begin{align} S(\ev_1) &= -\nabla_{\ev_1}\nn = -\frac{\partial\nn}{\partial\phi} = \cos\psi\sin\phi\,\xhat - \cos\psi\cos\phi\,\yhat = -\frac{\cos\psi}{R+\rho\cos\psi}\,\ev_1 \\ S(\ev_2) &= -\nabla_{\ev_2}\nn = -\frac{\partial\nn}{\partial\psi} = \sin\psi\cos\phi\,\xhat + \sin\psi\sin\phi\,\yhat - \cos\psi\,\zhat = -\frac{1}{\rho}\,\ev_2 . \end{align} Thus, $\ev_1$ and $\ev_2$ are principal directions, and the principal curvatures are $k_1=-\frac{\cos\psi}{R+\rho\cos\psi}$, $k_2=-\frac1\rho$.
Since the toroidal coordinates above are orthogonal, an orthonormal frame adapted to the torus is given by \begin{equation} \{\ee_1,\ee_2,\ee_3\} = \{\frac{\ev_1}{|\ev_1|},\frac{\ev_2}{|\ev_2|},\nn\} . \end{equation} Elsewhere, we have computed the basis of 1-forms dual to this frame, whose elements are \begin{align} \sigma_1 &= (R+\rho\cos\psi)\,d\phi ,\\ \sigma_2 &= \rho\,d\psi ,\\ \sigma_3 &= d\rho , \end{align} and the connection 1-forms, which are \begin{align} \omega_{12} &= \sin\psi\,d\phi = \frac{\sin\psi}{R+\rho\cos\psi}\,\sigma_1 ,\\ \omega_{13} &= -\cos\psi\,d\phi = -\frac{\cos\psi}{R+\rho\cos\psi}\,\sigma_1 ,\\ \omega_{23} &= -d\psi = -\frac1\rho\,\sigma_2 . \end{align} We can now determine the shape operator by taking derivatives of $\nn$, yielding \begin{align} S(\ee_1) &= -\nabla_{\ee_1}\ee_3 = -\omega_{3m}(\ee_1) \,\ee_m = -\frac{\cos\psi}{R+\rho\cos\psi}\sigma_1(\ee_1) \,\ee_1 = -\frac{\cos\psi}{R+\rho\cos\psi} \,\ee_1 ,\\ S(\ee_2) &= -\nabla_{\ee_2}\ee_3 = -\omega_{3m}(\ee_2) \,\ee_m = -\frac1\rho\sigma_2(\ee_2) \,\ee_2 = -\frac1\rho \,\ee_2 . \end{align} As before, $\ee_1$ and $\ee_2$ are principal directions, and the principal curvatures are $k_1=-\frac{\cos\psi}{R+\rho\cos\psi}$, $k_2=-\frac1\rho$.
We work entirely in the torus, with frame $\{\ee_1,\ee_2\}$ and dual basis $\{\sigma_1,\sigma_2\}=\{(R+\rho\cos\psi)\,d\phi,\rho\,d\psi\}$. The only independent connection 1-form is $\omega_{12}=\sin\psi\,d\phi$. We can no longer compute the shape operator $S$ or the principal curvatures $k_m$, but we can nonetheless determine the Gaussian curvature $K=k_1k_2$ using \begin{equation} d\omega_{12} = -K\,\sigma_1\wedge\sigma_2 , \end{equation} which yields \begin{equation} d\omega_{12} = \cos\psi\,d\psi\wedge d\phi = -\frac1\rho\frac{\cos\psi}{R+\rho\cos\psi} \sigma_1\wedge\sigma_2 , \end{equation} from which it follows that \begin{equation} K = \frac1\rho\frac{\cos\psi}{R+\rho\cos\psi} \end{equation}